This video is an Introductory lecture to Graph data structure.
This track of the course covers the topic "Graph".
In details, this track will cover:
Objective: The objective of this track is to familiarize the learners with Graphs.
Track Content:
Assessment: All Tracks in every week are associated with weekly contests.
This video is an Introductory lecture to Graph data structure.
This video introduces us to the Graph Representation through Adjacency Matrix.
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This video introduces us to the Graph Representation through Adjacency List.
This video discusses the implementation of the Adjacency List in C++.
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This video discusses the implementation of the Adjacency List in Java.
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This video discusses the comparison of the Adjacency Matrix vs List in C++.
This video discusses the Breadth First Search algorithm in the graph.
We discuss following versions of B.F.S. :
1. Given an undirected graph and a source vertex 's' ,print B.F.S. from given source.
2. B.F.S on disconnected graphs.
3. Print number of islands in a graph (or number of connected components in a graph).
This video familiarises us with the Applications of Breadth-First Search in the world of the computer programming.
This video discusses the Depth First Search algorithm in a graph.
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This video throws light on various applications of Depth First Search algorithm.
Given an Unweighted Graph and a source point, the task is to find the shortest path between the source point and every other point in the graph.
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This problem discusses the Detection of a Cycle in an Undirected Graph.
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This problem discusses the Detection of a Cycle in an Undirected Graph using DFS.
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This video discusses the problem of Topological Sorting using Kahn's BFS Based Algorithm.
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This problem discusses the Detection of a Cycle in an Undirected Graph using Kahn's algorithm.
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This video discusses the Topological Sorting using the DFS Based Algorithm.
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This video discusses the problem of finding the Shortest Path in Directed Acyclic Graph.
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This video discusses the concept of Minimum Spanning Tree and how can it be calculated using Prims Algorithm.
This video explains the concept and working of the Dijkstra Shortest path Algorithm through a few working examples.
Kosaraju's Algorithm Part 1
Kosaraju's Algorithm Part 2
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Explanation of the algorithm and working of it for negative weight cycles.
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This video introduces us to the Articulation Point or the cut vertices. We will also be looking at a few problems where we will need to find the articulation point for a connected and undirected graph.
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This video introduces us to the bridge in a graph. It discusses the solution of finding the bridges in an undirected graph.
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This video discusses the Tarjan's Algorithm for finding the strongly connected components in a directed graph.
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In this video Kruskal's algorithm and its implementation using Disjoint Set Data Structure is discussed
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Directed and Undirected Graphs
Representing Graphs
Following two are the most commonly used representations of a graph:
Adjacency list of vertex 0
head -> 1-> 4
Adjacency list of vertex 1
head -> 0-> 2-> 3-> 4
Adjacency list of vertex 2
head -> 1-> 3
Adjacency list of vertex 3
head -> 1-> 2-> 4
Adjacency list of vertex 4
head -> 0-> 1-> 3
Since the Queue is empty now, it means that the complete graph is traversed.
// C++ program to implement BFS traversal
// of a Graph
#include <bits/stdc++.h>
using namespace std;
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> adj[], int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
// Function to perform BFS traversal of the given Graph
void BFS(vector<int> adj[], int V)
{
// Initialize a boolean array
// to keep track of visited vertices
bool visited[V + 1];
// Mark all vertices not-visited initially
for (int i = 1; i <= V; i++)
visited[i] = false;
// Create a Queue to perform BFS
queue<int> q;
// Our source vertex is vertex
// numbered 1
int s = 1;
// Mark S visited and Push to queue
visited[s] = true;
q.push(s);
while (!q.empty()) {
// Pop element at front and print
int node = q.front();
q.pop();
cout << node << " ";
// Traverse the nodes adjacent to the currently
// poped element and push those elements to the
// queue which are not already visited
for (int i = 0; i < adj[node].size(); i++) {
if (visited[adj[node][i]] == false) {
// Mark it visited
visited[adj[node][i]] = true;
// Push it to the Queue
q.push(adj[node][i]);
}
}
}
}
// Driver code
int main()
{
int V = 6;
vector<int> adj[V + 1];
addEdge(adj, 1, 2);
addEdge(adj, 1, 3);
addEdge(adj, 2, 4);
addEdge(adj, 2, 5);
addEdge(adj, 3, 5);
addEdge(adj, 4, 5);
addEdge(adj, 4, 6);
addEdge(adj, 5, 6);
BFS(adj, V);
return 0;
}// Java code to illustrate BFS traversal
// in a Graph
import java.util.*;
class Graph {
// A utility function to add an edge in an
// undirected graph
static void addEdge(ArrayList<ArrayList<Integer> > adj,
int u, int v)
{
adj.get(u).add(v);
adj.get(v).add(u);
}
// Function to perform BFS traversal of a Graph
static void BFS(ArrayList<ArrayList<Integer> > adj, int V)
{
// Initialize a boolean array
// to keep track of visited vertices
boolean visited[] = new boolean[V+1];
// Mark all vertices not-visited initially
for (int i = 1; i <= V; i++)
visited[i] = false;
// Create a queue for BFS
LinkedList<Integer> queue = new LinkedList<Integer>();
// The start vertex or source vertex is 1
int s = 1;
// Mark the current node as
// visited and enqueue it
visited[s]=true;
queue.add(s);
while (queue.size() != 0)
{
// Dequeue a vertex from queue and print it
s = queue.poll();
System.out.print(s+" ");
// Traverse the nodes adjacent to the currently
// poped element and push those elements to the
// queue which are not already visited
for (int i = 0; i < adj.get(s).size(); i++) {
// Fetch adjacent node
int newNode = adj.get(s).get(i);
// Check if it is not visited
if(visited[newNode] == false)
{
// Mark it visited
visited[newNode] = true;
// Add it to queue
queue.add(newNode);
}
}
}
}
// Driver Code
public static void main(String[] args)
{
// Creating a graph with 6 vertices
int V = 6;
ArrayList<ArrayList<Integer> > adj
= new ArrayList<ArrayList<Integer> >(V+1);
for (int i = 0; i < V+1; i++)
adj.add(new ArrayList<Integer>());
// Adding edges one by one
addEdge(adj, 1, 2);
addEdge(adj, 1, 3);
addEdge(adj, 2, 4);
addEdge(adj, 2, 5);
addEdge(adj, 3, 5);
addEdge(adj, 4, 5);
addEdge(adj, 4, 6);
addEdge(adj, 5, 6);
BFS(adj, V);
}
}1 2 3 4 5 6
// C++ program to print DFS traversal from
// a given vertex in a given graph
#include<iostream>
#include<list>
using namespace std;
// Graph class represents a directed graph
// using adjacency list representation
class Graph
{
int V; // No. of vertices
// Pointer to an array containing
// adjacency lists
list<int> *adj;
// A recursive function used by DFS
void DFSUtil(int v, bool visited[]);
public:
Graph(int V); // Constructor
// function to add an edge to graph
void addEdge(int v, int w);
// DFS traversal of the vertices
// reachable from v
void DFS(int v);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::DFSUtil(int v, bool visited[])
{
// Mark the current node as visited and
// print it
visited[v] = true;
cout << v << " ";
// Recur for all the vertices adjacent
// to this vertex
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (!visited[*i])
DFSUtil(*i, visited);
}
// DFS traversal of the vertices reachable from v.
// It uses recursive DFSUtil()
void Graph::DFS(int v)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for (int i = 0; i < V; i++)
visited[i] = false;
// Call the recursive helper function
// to print DFS traversal
DFSUtil(v, visited);
}
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << "Following is Depth First Traversal"
" (starting from vertex 2) \n";
g.DFS(2);
return 0;
}// Java program to print DFS traversal from a given given graph
import java.io.*;
import java.util.*;
// This class represents a directed graph using adjacency list
// representation
class Graph
{
private int V; // No. of vertices
// Array of lists for Adjacency List Representation
private LinkedList<Integer> adj[];
// Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
//Function to add an edge into the graph
void addEdge(int v, int w)
{
adj[v].add(w); // Add w to v's list.
}
// A function used by DFS
void DFSUtil(int v,boolean visited[])
{
// Mark the current node as visited and print it
visited[v] = true;
System.out.print(v+" ");
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i = adj[v].listIterator();
while (i.hasNext())
{
int n = i.next();
if (!visited[n])
DFSUtil(n, visited);
}
}
// The function to do DFS traversal. It uses recursive DFSUtil()
void DFS(int v)
{
// Mark all the vertices as not visited(set as
// false by default in java)
boolean visited[] = new boolean[V];
// Call the recursive helper function to print DFS traversal
DFSUtil(v, visited);
}
public static void main(String args[])
{
Graph g = new Graph(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
System.out.println("Following is Depth First Traversal "+
"(starting from vertex 2)");
g.DFS(2);
}
}Following is Depth First Traversal (starting from vertex 2)
2 0 1 3
// A C++ Program to detect cycle in a graph
#include<iostream>
#include <list>
#include <limits.h>
using namespace std;
class Graph
{
int V; // No. of vertices
list<int> *adj; // Pointer to an array containing adjacency lists
bool isCyclicUtil(int v, bool visited[], bool *rs); // used by isCyclic()
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // to add an edge to graph
bool isCyclic(); // returns true if there is a cycle in this graph
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
// Utility function to detect cycle in a Graph
bool Graph::isCyclicUtil(int v, bool visited[], bool *recStack)
{
if(visited[v] == false)
{
// Mark the current node as visited and part of recursion stack
visited[v] = true;
recStack[v] = true;
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i)
{
if ( !visited[*i] && isCyclicUtil(*i, visited, recStack) )
return true;
else if (recStack[*i])
return true;
}
}
recStack[v] = false; // remove the vertex from recursion stack
return false;
}
// Returns true if the graph contains a cycle, else false
bool Graph::isCyclic()
{
// Mark all the vertices as not visited and
// not part of recursion stack
bool *visited = new bool[V];
bool *recStack = new bool[V];
for(int i = 0; i < V; i++)
{
visited[i] = false;
recStack[i] = false;
}
// Call the recursive helper function to detect
// cycle in different DFS trees
for(int i = 0; i < V; i++)
if (isCyclicUtil(i, visited, recStack))
return true;
return false;
}
// Driver Code
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
if(g.isCyclic())
cout << "Graph contains cycle";
else
cout << "Graph doesn't contain cycle";
return 0;
}
// A Java Program to detect cycle in a graph
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
class Graph {
private final int V;
private final List<List<Integer>> adj;
public Graph(int V)
{
this.V = V;
adj = new ArrayList<>(V);
for (int i = 0; i < V; i++)
adj.add(new LinkedList<>());
}
// Utility Function to check cycle in a Graph
private boolean isCyclicUtil(int i, boolean[] visited,
boolean[] recStack)
{
// Mark the current node as visited and
// part of recursion stack
if (recStack[i])
return true;
if (visited[i])
return false;
visited[i] = true;
recStack[i] = true;
List<Integer> children = adj.get(i);
for (Integer c: children)
if (isCyclicUtil(c, visited, recStack))
return true;
recStack[i] = false;
return false;
}
private void addEdge(int source, int dest) {
adj.get(source).add(dest);
}
// Returns true if the graph contains a
// cycle, else false.
// This function is a variation of DFS() in
// https://www.cdn.geeksforgeeks.org/archives/18212
private boolean isCyclic()
{
// Mark all the vertices as not visited and
// not part of recursion stack
boolean[] visited = new boolean[V];
boolean[] recStack = new boolean[V];
// Call the recursive helper function to
// detect cycle in different DFS trees
for (int i = 0; i < V; i++)
if (isCyclicUtil(i, visited, recStack))
return true;
return false;
}
// Driver code
public static void main(String[] args)
{
Graph graph = new Graph(4);
graph.addEdge(0, 1);
graph.addEdge(0, 2);
graph.addEdge(1, 2);
graph.addEdge(2, 0);
graph.addEdge(2, 3);
graph.addEdge(3, 3);
if(graph.isCyclic())
System.out.println("Graph contains cycle");
else
System.out.println("Graph doesn't "
+ "contain cycle");
}
}
Graph contains cycle
Given a graph and a source vertex in the graph, find the shortest paths from source to all vertices in the given graph.
1) Initialize distances of all vertices as infinite.
2) Create an empty priority_queue pq. Every item
of pq is a pair (weight, vertex). Weight (or
distance) is used as the first item of pair
as the first item is by default used to compare
two pairs
3) Insert source vertex into pq and make its
distance as 0.
4) While either pq doesn't become empty
a) Extract minimum distance vertex from pq.
Let the extracted vertex be u.
b) Loop through all adjacent of u and do
following for every vertex v.
// If there is a shorter path to v
// through u.
If dist[v] > dist[u] + weight(u, v)
(i) Update distance of v, i.e., do
dist[v] = dist[u] + weight(u, v)
(ii) Insert v into the pq (Even if v is
already there)
5) Print distance array dist[] to print all shortest
paths.
// Program to find Dijkstra's shortest path using
// min heap in STL
#include<bits/stdc++.h>
using namespace std;
# define INF 0x3f3f3f3f
// iPair ==> Integer Pair
typedef pair<int, int> iPair;
// To add an edge
void addEdge(vector <pair<int, int> > adj[], int u,
int v, int wt)
{
adj[u].push_back(make_pair(v, wt));
adj[v].push_back(make_pair(u, wt));
}
// Prints distance of shortest paths from the source
// vertex to all other vertices
void shortestPath(vector<pair<int,int> > adj[], int V, int src)
{
// Create a priority queue to store vertices that
// are being preprocessed. This is weird syntax in C++.
// Refer below link for details of this syntax
// http://geeksquiz.com/implement-min-heap-using-stl/
priority_queue< iPair, vector <iPair> , greater<iPair> > pq;
// Create a vector for distances and initialize all
// distances as infinite (INF)
vector<int> dist(V, INF);
// Insert source itself in priority queue and initialize
// its distance as 0.
pq.push(make_pair(0, src));
dist[src] = 0;
/* Looping till priority queue becomes empty (or all
distances are not finalized) */
while (!pq.empty())
{
// The first vertex in pair is the minimum distance
// vertex, extract it from priority queue.
// vertex label is stored in second of pair (it
// has to be done this way to keep the vertices
// sorted distance (distance must be first item
// in pair)
int u = pq.top().second;
pq.pop();
// Get all adjacent of u.
for (auto x : adj[u])
{
// Get vertex label and weight of current adjacent
// of u.
int v = x.first;
int weight = x.second;
// If there is shorted path to v through u.
if (dist[v] > dist[u] + weight)
{
// Updating distance of v
dist[v] = dist[u] + weight;
pq.push(make_pair(dist[v], v));
}
}
}
// Print shortest distances stored in dist[]
printf("Vertex Distance from Source\n");
for (int i = 0; i < V; ++i)
printf("%d \t\t %d\n", i, dist[i]);
}
// Driver Code
int main()
{
int V = 9;
vector<iPair > adj[V];
// making above shown graph
addEdge(adj, 0, 1, 4);
addEdge(adj, 0, 7, 8);
addEdge(adj, 1, 2, 8);
addEdge(adj, 1, 7, 11);
addEdge(adj, 2, 3, 7);
addEdge(adj, 2, 8, 2);
addEdge(adj, 2, 5, 4);
addEdge(adj, 3, 4, 9);
addEdge(adj, 3, 5, 14);
addEdge(adj, 4, 5, 10);
addEdge(adj, 5, 6, 2);
addEdge(adj, 6, 7, 1);
addEdge(adj, 6, 8, 6);
addEdge(adj, 7, 8, 7);
shortestPath(adj, V, 0);
return 0;
}
// A Java program for Dijkstra's single source shortest path algorithm.
// The program is for adjacency matrix representation of the graph
import java.util.*;
import java.lang.*;
import java.io.*;
class ShortestPath {
// A utility function to find the vertex with minimum distance value,
// from the set of vertices not yet included in shortest path tree
static final int V = 9;
int minDistance(int dist[], Boolean sptSet[])
{
// Initialize min value
int min = Integer.MAX_VALUE, min_index = -1;
for (int v = 0; v < V; v++)
if (sptSet[v] == false && dist[v] <= min) {
min = dist[v];
min_index = v;
}
return min_index;
}
// A utility function to print the constructed distance array
void printSolution(int dist[], int n)
{
System.out.println("Vertex Distance from Source\n");
for (int i = 0; i < V; i++)
System.out.println(i + " " + dist[i]+"\n");
}
// Function that implements Dijkstra's single source shortest path
// algorithm for a graph represented using adjacency matrix
// representation
void dijkstra(int graph[][], int src)
{
int dist[] = new int[V]; // The output array. dist[i] will hold
// the shortest distance from src to i
// sptSet[i] will true if vertex i is included in shortest
// path tree or shortest distance from src to i is finalized
Boolean sptSet[] = new Boolean[V];
// Initialize all distances as INFINITE and stpSet[] as false
for (int i = 0; i < V; i++) {
dist[i] = Integer.MAX_VALUE;
sptSet[i] = false;
}
// Distance of source vertex from itself is always 0
dist[src] = 0;
// Find shortest path for all vertices
for (int count = 0; count < V - 1; count++) {
// Pick the minimum distance vertex from the set of vertices
// not yet processed. u is always equal to src in first
// iteration.
int u = minDistance(dist, sptSet);
// Mark the picked vertex as processed
sptSet[u] = true;
// Update dist value of the adjacent vertices of the
// picked vertex.
for (int v = 0; v < V; v++)
// Update dist[v] only if is not in sptSet, there is an
// edge from u to v, and total weight of path from src to
// v through u is smaller than current value of dist[v]
if (!sptSet[v] && graph[u][v] != 0 &&
dist[u] != Integer.MAX_VALUE && dist[u] + graph[u][v] < dist[v])
dist[v] = dist[u] + graph[u][v];
}
// print the constructed distance array
printSolution(dist, V);
}
// Driver method
public static void main(String[] args)
{
/* Let us create the example graph discussed above */
int graph[][] = new int[][] { { 0, 4, 0, 0, 0, 0, 0, 8, 0 },
{ 4, 0, 8, 0, 0, 0, 0, 11, 0 },
{ 0, 8, 0, 7, 0, 4, 0, 0, 2 },
{ 0, 0, 7, 0, 9, 14, 0, 0, 0 },
{ 0, 0, 0, 9, 0, 10, 0, 0, 0 },
{ 0, 0, 4, 14, 10, 0, 2, 0, 0 },
{ 0, 0, 0, 0, 0, 2, 0, 1, 6 },
{ 8, 11, 0, 0, 0, 0, 1, 0, 7 },
{ 0, 0, 2, 0, 0, 0, 6, 7, 0 } };
ShortestPath t = new ShortestPath();
t.dijkstra(graph, 0);
}
}
Vertex Distance from Source
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
// A C++ program for Bellman-Ford's single source
// shortest path algorithm.
#include <bits/stdc++.h>
// a structure to represent a weighted edge in graph
struct Edge
{
int src, dest, weight;
};
// a structure to represent a connected, directed and
// weighted graph
struct Graph
{
// V-> Number of vertices, E-> Number of edges
int V, E;
// graph is represented as an array of edges.
struct Edge* edge;
};
// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E)
{
struct Graph* graph = new Graph;
graph->V = V;
graph->E = E;
graph->edge = new Edge[E];
return graph;
}
// A utility function used to print the solution
void printArr(int dist[], int n)
{
printf("Vertex Distance from Source\n");
for (int i = 0; i < n; ++i)
printf("%d \t\t %d\n", i, dist[i]);
}
// The main function that finds shortest distances from src to
// all other vertices using Bellman-Ford algorithm. The function
// also detects negative weight cycle
void BellmanFord(struct Graph* graph, int src)
{
int V = graph->V;
int E = graph->E;
int dist[V];
// Step 1: Initialize distances from src to all other vertices
// as INFINITE
for (int i = 0; i < V; i++)
dist[i] = INT_MAX;
dist[src] = 0;
// Step 2: Relax all edges |V| - 1 times. A simple shortest
// path from src to any other vertex can have at-most |V| - 1
// edges
for (int i = 1; i <= V-1; i++)
{
for (int j = 0; j < E; j++)
{
int u = graph->edge[j].src;
int v = graph->edge[j].dest;
int weight = graph->edge[j].weight;
if (dist[u] != INT_MAX && dist[u] + weight < dist[v])
dist[v] = dist[u] + weight;
}
}
// Step 3: check for negative-weight cycles. The above step
// guarantees shortest distances if graph doesn't contain
// negative weight cycle. If we get a shorter path, then there
// is a cycle.
for (int i = 0; i < E; i++)
{
int u = graph->edge[i].src;
int v = graph->edge[i].dest;
int weight = graph->edge[i].weight;
if (dist[u] != INT_MAX && dist[u] + weight < dist[v])
printf("Graph contains negative weight cycle");
}
printArr(dist, V);
return;
}
// Driver program to test above functions
int main()
{
/* Let us create the graph given in above example */
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
struct Graph* graph = createGraph(V, E);
// add edge 0-1 (or A-B in above figure)
graph->edge[0].src = 0;
graph->edge[0].dest = 1;
graph->edge[0].weight = -1;
// add edge 0-2 (or A-C in above figure)
graph->edge[1].src = 0;
graph->edge[1].dest = 2;
graph->edge[1].weight = 4;
// add edge 1-2 (or B-C in above figure)
graph->edge[2].src = 1;
graph->edge[2].dest = 2;
graph->edge[2].weight = 3;
// add edge 1-3 (or B-D in above figure)
graph->edge[3].src = 1;
graph->edge[3].dest = 3;
graph->edge[3].weight = 2;
// add edge 1-4 (or A-E in above figure)
graph->edge[4].src = 1;
graph->edge[4].dest = 4;
graph->edge[4].weight = 2;
// add edge 3-2 (or D-C in above figure)
graph->edge[5].src = 3;
graph->edge[5].dest = 2;
graph->edge[5].weight = 5;
// add edge 3-1 (or D-B in above figure)
graph->edge[6].src = 3;
graph->edge[6].dest = 1;
graph->edge[6].weight = 1;
// add edge 4-3 (or E-D in above figure)
graph->edge[7].src = 4;
graph->edge[7].dest = 3;
graph->edge[7].weight = -3;
BellmanFord(graph, 0);
return 0;
}// A Java program for Bellman-Ford's single source shortest path
// algorithm.
import java.util.*;
import java.lang.*;
import java.io.*;
// A class to represent a connected, directed and weighted graph
class Graph
{
// A class to represent a weighted edge in graph
class Edge {
int src, dest, weight;
Edge() {
src = dest = weight = 0;
}
};
int V, E;
Edge edge[];
// Creates a graph with V vertices and E edges
Graph(int v, int e)
{
V = v;
E = e;
edge = new Edge[e];
for (int i=0; i<e; ++i)
edge[i] = new Edge();
}
// The main function that finds shortest distances from src
// to all other vertices using Bellman-Ford algorithm. The
// function also detects negative weight cycle
void BellmanFord(Graph graph,int src)
{
int V = graph.V, E = graph.E;
int dist[] = new int[V];
// Step 1: Initialize distances from src to all other
// vertices as INFINITE
for (int i=0; i<V; ++i)
dist[i] = Integer.MAX_VALUE;
dist[src] = 0;
// Step 2: Relax all edges |V| - 1 times. A simple
// shortest path from src to any other vertex can
// have at-most |V| - 1 edges
for (int i=1; i<V; ++i)
{
for (int j=0; j<E; ++j)
{
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u]!=Integer.MAX_VALUE &&
dist[u]+weight<dist[v])
dist[v]=dist[u]+weight;
}
}
// Step 3: check for negative-weight cycles. The above
// step guarantees shortest distances if graph doesn't
// contain negative weight cycle. If we get a shorter
// path, then there is a cycle.
for (int j=0; j<E; ++j)
{
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u] != Integer.MAX_VALUE &&
dist[u]+weight < dist[v])
System.out.println("Graph contains negative weight cycle");
}
printArr(dist, V);
}
// A utility function used to print the solution
void printArr(int dist[], int V)
{
System.out.println("Vertex Distance from Source");
for (int i=0; i<V; ++i)
System.out.println(i+"\t\t"+dist[i]);
}
// Driver method to test above function
public static void main(String[] args)
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
Graph graph = new Graph(V, E);
// add edge 0-1 (or A-B in above figure)
graph.edge[0].src = 0;
graph.edge[0].dest = 1;
graph.edge[0].weight = -1;
// add edge 0-2 (or A-C in above figure)
graph.edge[1].src = 0;
graph.edge[1].dest = 2;
graph.edge[1].weight = 4;
// add edge 1-2 (or B-C in above figure)
graph.edge[2].src = 1;
graph.edge[2].dest = 2;
graph.edge[2].weight = 3;
// add edge 1-3 (or B-D in above figure)
graph.edge[3].src = 1;
graph.edge[3].dest = 3;
graph.edge[3].weight = 2;
// add edge 1-4 (or A-E in above figure)
graph.edge[4].src = 1;
graph.edge[4].dest = 4;
graph.edge[4].weight = 2;
// add edge 3-2 (or D-C in above figure)
graph.edge[5].src = 3;
graph.edge[5].dest = 2;
graph.edge[5].weight = 5;
// add edge 3-1 (or D-B in above figure)
graph.edge[6].src = 3;
graph.edge[6].dest = 1;
graph.edge[6].weight = 1;
// add edge 4-3 (or E-D in above figure)
graph.edge[7].src = 4;
graph.edge[7].dest = 3;
graph.edge[7].weight = -3;
graph.BellmanFord(graph, 0);
}
}Vertex Distance from Source
0 0
1 -1
2 2
3 -2
4 1
1) Initialize all vertices as not visited.
2) Do following for every vertex 'v'.
(a) If 'v' is not visited before, call DFSUtil(v)
(b) Print new line character
// This Function performs DFS traversal
// of vertex v.
DFSUtil(v)
1) Mark 'v' as visited.
2) Print 'v'
3) Do following for every adjacent 'u' of 'v'.
If 'u' is not visited, then recursively call DFSUtil(u)
// C++ program to print connected components in
// an undirected graph
#include <iostream>
#include <list>
using namespace std;
// Graph class represents a undirected graph
// using adjacency list representation
class Graph {
int V; // No. of vertices
// Pointer to an array containing adjacency lists
list<int>* adj;
// A function used by DFS
void DFSUtil(int v, bool visited[]);
public:
Graph(int V); // Constructor
void addEdge(int v, int w);
void connectedComponents();
};
// Method to print connected components in an
// undirected graph
void Graph::connectedComponents()
{
// Mark all the vertices as not visited
bool* visited = new bool[V];
for (int v = 0; v < V; v++)
visited[v] = false;
for (int v = 0; v < V; v++) {
if (visited[v] == false) {
// print all reachable vertices
// from v
DFSUtil(v, visited);
cout << "\n";
}
}
}
void Graph::DFSUtil(int v, bool visited[])
{
// Mark the current node as visited and print it
visited[v] = true;
cout << v << " ";
// Recur for all the vertices
// adjacent to this vertex
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (!visited[*i])
DFSUtil(*i, visited);
}
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
// method to add an undirected edge
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
// Drive program to test above
int main()
{
// Create a graph given in the above diagram
Graph g(5); // 5 vertices numbered from 0 to 4
g.addEdge(1, 0);
g.addEdge(1, 2);
g.addEdge(3, 4);
cout << "Following are connected components \n";
g.connectedComponents();
return 0;
}// Java program to print connected components in
// an undirected graph
import java.util.LinkedList;
class Graph {
// A user define class to represent a graph.
// A graph is an array of adjacency lists.
// Size of array will be V (number of vertices
// in graph)
int V;
LinkedList<Integer>[] adjListArray;
// constructor
Graph(int V)
{
this.V = V;
// define the size of array as
// number of vertices
adjListArray = new LinkedList[V];
// Create a new list for each vertex
// such that adjacent nodes can be stored
for (int i = 0; i < V; i++) {
adjListArray[i] = new LinkedList<Integer>();
}
}
// Adds an edge to an undirected graph
void addEdge(int src, int dest)
{
// Add an edge from src to dest.
adjListArray[src].add(dest);
// Since graph is undirected, add an edge from dest
// to src also
adjListArray[dest].add(src);
}
void DFSUtil(int v, boolean[] visited)
{
// Mark the current node as visited and print it
visited[v] = true;
System.out.print(v + " ");
// Recur for all the vertices
// adjacent to this vertex
for (int x : adjListArray[v]) {
if (!visited[x])
DFSUtil(x, visited);
}
}
void connectedComponents()
{
// Mark all the vertices as not visited
boolean[] visited = new boolean[V];
for (int v = 0; v < V; ++v) {
if (!visited[v]) {
// print all reachable vertices
// from v
DFSUtil(v, visited);
System.out.println();
}
}
}
// Driver program to test above
public static void main(String[] args)
{
// Create a graph given in the above diagram
Graph g = new Graph(5); // 5 vertices numbered from 0 to 4
g.addEdge(1, 0);
g.addEdge(1, 2);
g.addEdge(3, 4);
System.out.println("Following are connected components");
g.connectedComponents();
}
}Following are connected components
0 1 2
3 4
What is Minimum Spanning Tree?
Given a connected and undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together. A single graph can have many different spanning trees. A minimum spanning tree (MST) or minimum weight spanning tree for a weighted, connected and undirected graph is a spanning tree with weight less than or equal to the weight of every other spanning tree. The weight of a spanning tree is the sum of weights given to each edge of the spanning tree.Prim's Algorithm
Prim’s algorithm is also a Greedy algorithm. It starts with an empty spanning tree. The idea is to maintain two sets of vertices. The first set contains the vertices already included in the MST, the other set contains the vertices not yet included. At every step, it considers all the edges that connect the two sets, and picks the minimum weight edge from these edges. After picking the edge, it moves the other endpoint of the edge to the set containing MST.
How to implement the above algorithm?
We use a boolean array mstSet[] to represent the set of vertices included in MST. If a value mstSet[v] is true, then vertex v is included in MST, otherwise not. Array key[] is used to store key values of all vertices. Another array parent[] to store indexes of parent nodes in MST. The parent array is the output array which is used to show the constructed MST.// A C++ program for Prim's Minimum
// Spanning Tree (MST) algorithm. The program is
// for adjacency matrix representation of the graph
#include <bits/stdc++.h>
using namespace std;
// Number of vertices in the graph
#define V 5
// A utility function to find the vertex with
// minimum key value, from the set of vertices
// not yet included in MST
int minKey(int key[], bool mstSet[])
{
// Initialize min value
int min = INT_MAX, min_index;
for (int v = 0; v < V; v++)
if (mstSet[v] == false && key[v] < min)
min = key[v], min_index = v;
return min_index;
}
// A utility function to print the
// constructed MST stored in parent[]
int printMST(int parent[], int graph[V][V])
{
cout<<"Edge \tWeight\n";
for (int i = 1; i < V; i++)
cout<<parent[i]<<" - "<<i<<" \t"<<graph[i][parent[i]]<<" \n";
}
// Function to construct and print MST for
// a graph represented using adjacency
// matrix representation
void primMST(int graph[V][V])
{
// Array to store constructed MST
int parent[V];
// Key values used to pick minimum weight edge in cut
int key[V];
// To represent set of vertices not yet included in MST
bool mstSet[V];
// Initialize all keys as INFINITE
for (int i = 0; i < V; i++)
key[i] = INT_MAX, mstSet[i] = false;
// Always include first 1st vertex in MST.
// Make key 0 so that this vertex is picked as first vertex.
key[0] = 0;
parent[0] = -1; // First node is always root of MST
// The MST will have V vertices
for (int count = 0; count < V - 1; count++)
{
// Pick the minimum key vertex from the
// set of vertices not yet included in MST
int u = minKey(key, mstSet);
// Add the picked vertex to the MST Set
mstSet[u] = true;
// Update key value and parent index of
// the adjacent vertices of the picked vertex.
// Consider only those vertices which are not
// yet included in MST
for (int v = 0; v < V; v++)
// graph[u][v] is non zero only for adjacent vertices of m
// mstSet[v] is false for vertices not yet included in MST
// Update the key only if graph[u][v] is smaller than key[v]
if (graph[u][v] && mstSet[v] == false && graph[u][v] < key[v])
parent[v] = u, key[v] = graph[u][v];
}
// print the constructed MST
printMST(parent, graph);
}
// Driver code
int main()
{
/* Let us create the following graph
2 3
(0)--(1)--(2)
| / \ |
6| 8/ \5 |7
| / \ |
(3)-------(4)
9 */
int graph[V][V] = { { 0, 2, 0, 6, 0 },
{ 2, 0, 3, 8, 5 },
{ 0, 3, 0, 0, 7 },
{ 6, 8, 0, 0, 9 },
{ 0, 5, 7, 9, 0 } };
// Print the solution
primMST(graph);
return 0;
}
// A Java program for Prim's Minimum Spanning Tree (MST) algorithm.
// The program is for adjacency matrix representation of the graph
import java.util.*;
import java.lang.*;
import java.io.*;
class MST {
// Number of vertices in the graph
private static final int V = 5;
// A utility function to find the vertex with minimum key
// value, from the set of vertices not yet included in MST
int minKey(int key[], Boolean mstSet[])
{
// Initialize min value
int min = Integer.MAX_VALUE, min_index = -1;
for (int v = 0; v < V; v++)
if (mstSet[v] == false && key[v] < min) {
min = key[v];
min_index = v;
}
return min_index;
}
// A utility function to print the constructed MST stored in
// parent[]
void printMST(int parent[], int graph[][])
{
System.out.println("Edge \tWeight");
for (int i = 1; i < V; i++)
System.out.println(parent[i] + " - " + i + "\t" + graph[i][parent[i]]);
}
// Function to construct and print MST for a graph represented
// using adjacency matrix representation
void primMST(int graph[][])
{
// Array to store constructed MST
int parent[] = new int[V];
// Key values used to pick minimum weight edge in cut
int key[] = new int[V];
// To represent set of vertices not yet included in MST
Boolean mstSet[] = new Boolean[V];
// Initialize all keys as INFINITE
for (int i = 0; i < V; i++) {
key[i] = Integer.MAX_VALUE;
mstSet[i] = false;
}
// Always include first 1st vertex in MST.
key[0] = 0; // Make key 0 so that this vertex is
// picked as first vertex
parent[0] = -1; // First node is always root of MST
// The MST will have V vertices
for (int count = 0; count < V - 1; count++) {
// Pick thd minimum key vertex from the set of vertices
// not yet included in MST
int u = minKey(key, mstSet);
// Add the picked vertex to the MST Set
mstSet[u] = true;
// Update key value and parent index of the adjacent
// vertices of the picked vertex. Consider only those
// vertices which are not yet included in MST
for (int v = 0; v < V; v++)
// graph[u][v] is non zero only for adjacent vertices of m
// mstSet[v] is false for vertices not yet included in MST
// Update the key only if graph[u][v] is smaller than key[v]
if (graph[u][v] != 0 && mstSet[v] == false && graph[u][v] < key[v]) {
parent[v] = u;
key[v] = graph[u][v];
}
}
// print the constructed MST
printMST(parent, graph);
}
public static void main(String[] args)
{
/* Let us create the following graph
2 3
(0)--(1)--(2)
| / \ |
6| 8/ \5 |7
| / \ |
(3)-------(4)
9 */
MST t = new MST();
int graph[][] = new int[][] { { 0, 2, 0, 6, 0 },
{ 2, 0, 3, 8, 5 },
{ 0, 3, 0, 0, 7 },
{ 6, 8, 0, 0, 9 },
{ 0, 5, 7, 9, 0 } };
// Print the solution
t.primMST(graph);
}
}
Edge Weight
0 - 1 2
1 - 2 3
0 - 3 6
1 - 4 5
How to find all articulation points in a given graph?
A simple approach is to one by one remove all vertices and see if removal of a vertex causes disconnected graph. Following are steps of a simple approach for the connected graph.
low[u] = min(disc[u], disc[w])
where w is an ancestor of u and there is a back edge from
some descendant of u to w.
// A C++ program to find articulation points
// in an undirected graph
#include<iostream>
#include <list>
#define NIL -1
using namespace std;
// A class that represents an undirected graph
class Graph
{
int V; // No. of vertices
list<int> *adj; // A dynamic array of adjacency lists
void APUtil(int v, bool visited[], int disc[], int low[],
int parent[], bool ap[]);
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void AP(); // prints articulation points
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v); // Note: the graph is undirected
}
// A recursive function that find articulation points using DFS traversal
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
// ap[] --> Store articulation points
void Graph::APUtil(int u, bool visited[], int disc[],
int low[], int parent[], bool ap[])
{
// A static variable is used for simplicity,
// we can avoid the use of static
// variable by passing a pointer
static int time = 0;
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = *i; // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
APUtil(v, visited, disc, low, parent, ap);
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more chilren.
if (parent[u] == NIL && children > 1)
ap[u] = true;
// (2) If u is not root and low value of one of its child is more
// than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
ap[u] = true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = min(low[u], disc[v]);
}
}
// The function to do DFS traversal.
// It uses recursive function APUtil()
void Graph::AP()
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
bool *ap = new bool[V]; // To store articulation points
// Initialize parent and visited, and ap(articulation point) arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
ap[i] = false;
}
// Call the recursive helper function to find articulation points
// in DFS tree rooted with vertex 'i'
for (int i = 0; i < V; i++)
if (visited[i] == false)
APUtil(i, visited, disc, low, parent, ap);
// Now ap[] contains articulation points, print them
for (int i = 0; i < V; i++)
if (ap[i] == true)
cout << i << " ";
}
// Driver program to test above function
int main()
{
// Create graphs given in above diagrams
cout << "\nArticulation points in first graph \n";
Graph g1(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.AP();
cout << "\nArticulation points in second graph \n";
Graph g2(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.AP();
cout << "\nArticulation points in third graph \n";
Graph g3(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.AP();
return 0;
}
// A Java program to find articulation points
// in an undirected graph
import java.io.*;
import java.util.*;
import java.util.LinkedList;
// This class represents an undirected graph using
// adjacency list representation
class Graph
{
private int V; // No. of vertices
// Array of lists for Adjacency List Representation
private LinkedList<Integer> adj[];
int time = 0;
static final int NIL = -1;
// Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
//Function to add an edge into the graph
void addEdge(int v, int w)
{
adj[v].add(w); // Add w to v's list.
adj[w].add(v); //Add v to w's list
}
// A recursive function that find articulation points using DFS
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
// ap[] --> Store articulation points
void APUtil(int u, boolean visited[], int disc[],
int low[], int parent[], boolean ap[])
{
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
Iterator<Integer> i = adj[u].iterator();
while (i.hasNext())
{
int v = i.next(); // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
APUtil(v, visited, disc, low, parent, ap);
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = Math.min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more chilren.
if (parent[u] == NIL && children > 1)
ap[u] = true;
// (2) If u is not root and low value of one of its child
// is more than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
ap[u] = true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = Math.min(low[u], disc[v]);
}
}
// The function to do DFS traversal.
// It uses recursive function APUtil()
void AP()
{
// Mark all the vertices as not visited
boolean visited[] = new boolean[V];
int disc[] = new int[V];
int low[] = new int[V];
int parent[] = new int[V];
boolean ap[] = new boolean[V]; // To store articulation points
// Initialize parent and visited, and ap(articulation point)
// arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
ap[i] = false;
}
// Call the recursive helper function to find articulation
// points in DFS tree rooted with vertex 'i'
for (int i = 0; i < V; i++)
if (visited[i] == false)
APUtil(i, visited, disc, low, parent, ap);
// Now ap[] contains articulation points, print them
for (int i = 0; i < V; i++)
if (ap[i] == true)
System.out.print(i+" ");
}
// Driver method
public static void main(String args[])
{
// Create graphs given in above diagrams
System.out.println("Articulation points in first graph ");
Graph g1 = new Graph(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.AP();
System.out.println();
System.out.println("Articulation points in Second graph");
Graph g2 = new Graph(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.AP();
System.out.println();
System.out.println("Articulation points in Third graph ");
Graph g3 = new Graph(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.AP();
}
}Articulation points in first graph
0 3
Articulation points in second graph
1 2
Articulation points in third graph
1
How does this work?
The above algorithm is DFS based. It does DFS two times. DFS of a graph produces a single tree if all vertices are reachable from the DFS starting point. Otherwise, DFS produces a forest. So DFS of a graph with only one SCC always produces a tree. The important point to note is DFS may produce a tree or a forest when there are more than one SCCs depending upon the chosen starting point. For example, in the above diagram, if we start DFS from vertices 0 or 1 or 2, we get a tree as output. And if we start at 3 or 4, we get a forest. To find and print all SCCs, we would want to start DFS from vertex 4 (which is a sink vertex), then move to 3 which is sunk in the remaining set (set excluding 4) and finally any of the remaining vertices (0, 1, 2). So how do we find this sequence of picking vertices as starting points of DFS? Unfortunately, there is no direct way of getting this sequence. However, if we do a DFS of graph and store vertices according to their finish times, we make sure that the finish time of a vertex that connects to other SCCs (other than its own SCC), will always be greater than finish time of vertices in the other SCC (See this for proof). For example, in DFS of above example graph, the finish time of 0 is always greater than 3 and 4 (irrespective of the sequence of vertices considered for DFS). And the finish time of 3 is always greater than 4. DFS doesn't guarantee about other vertices, for example, finish times of 1 and 2 may be smaller or greater than 3 and 4 depending upon the sequence of vertices considered for DFS. So to use this property, we do DFS traversal of the complete graph and push every finished vertex to a stack. In stack, 3 always appears after 4, and 0 appear after both 3 and 4.
// C++ Implementation of Kosaraju's algorithm to print all SCCs
#include <iostream>
#include <list>
#include <stack>
using namespace std;
class Graph
{
int V; // No. of vertices
list<int> *adj; // An array of adjacency lists
// Fills Stack with vertices (in increasing order of finishing
// times). The top element of stack has the maximum finishing
// time
void fillOrder(int v, bool visited[], stack<int> &Stack);
// A recursive function to print DFS starting from v
void DFSUtil(int v, bool visited[]);
public:
Graph(int V);
void addEdge(int v, int w);
// The main function that finds and prints strongly connected
// components
void printSCCs();
// Function that returns reverse (or transpose) of this graph
Graph getTranspose();
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
// A recursive function to print DFS starting from v
void Graph::DFSUtil(int v, bool visited[])
{
// Mark the current node as visited and print it
visited[v] = true;
cout << v << " ";
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (!visited[*i])
DFSUtil(*i, visited);
}
Graph Graph::getTranspose()
{
Graph g(V);
for (int v = 0; v < V; v++)
{
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i)
{
g.adj[*i].push_back(v);
}
}
return g;
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::fillOrder(int v, bool visited[], stack<int> &Stack)
{
// Mark the current node as visited and print it
visited[v] = true;
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i)
if(!visited[*i])
fillOrder(*i, visited, Stack);
// All vertices reachable from v are processed by now, push v
Stack.push(v);
}
// The main function that finds and prints all
// strongly connected components
void Graph::printSCCs()
{
stack<int> Stack;
// Mark all the vertices as not visited (For first DFS)
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Fill vertices in stack according to their finishing times
for(int i = 0; i < V; i++)
if(visited[i] == false)
fillOrder(i, visited, Stack);
// Create a reversed graph
Graph gr = getTranspose();
// Mark all the vertices as not visited (For second DFS)
for(int i = 0; i < V; i++)
visited[i] = false;
// Now process all vertices in order defined by Stack
while (Stack.empty() == false)
{
// Pop a vertex from stack
int v = Stack.top();
Stack.pop();
// Print Strongly connected component of the popped vertex
if (visited[v] == false)
{
gr.DFSUtil(v, visited);
cout << endl;
}
}
}
// Driver program to test above functions
int main()
{
// Create a graph given in the above diagram
Graph g(5);
g.addEdge(1, 0);
g.addEdge(0, 2);
g.addEdge(2, 1);
g.addEdge(0, 3);
g.addEdge(3, 4);
cout << "Following are strongly connected components in "
"given graph \n";
g.printSCCs();
return 0;
}// Java implementation of Kosaraju's algorithm to print all SCCs
import java.io.*;
import java.util.*;
import java.util.LinkedList;
// This class represents a directed graph using adjacency list
// representation
class Graph
{
private int V; // No. of vertices
private LinkedList<Integer> adj[]; //Adjacency List
//Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
//Function to add an edge into the graph
void addEdge(int v, int w) { adj[v].add(w); }
// A recursive function to print DFS starting from v
void DFSUtil(int v,boolean visited[])
{
// Mark the current node as visited and print it
visited[v] = true;
System.out.print(v + " ");
int n;
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i =adj[v].iterator();
while (i.hasNext())
{
n = i.next();
if (!visited[n])
DFSUtil(n,visited);
}
}
// Function that returns reverse (or transpose) of this graph
Graph getTranspose()
{
Graph g = new Graph(V);
for (int v = 0; v < V; v++)
{
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i =adj[v].listIterator();
while(i.hasNext())
g.adj[i.next()].add(v);
}
return g;
}
void fillOrder(int v, boolean visited[], Stack stack)
{
// Mark the current node as visited and print it
visited[v] = true;
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i = adj[v].iterator();
while (i.hasNext())
{
int n = i.next();
if(!visited[n])
fillOrder(n, visited, stack);
}
// All vertices reachable from v are processed by now,
// push v to Stack
stack.push(new Integer(v));
}
// The main function that finds and prints all strongly
// connected components
void printSCCs()
{
Stack stack = new Stack();
// Mark all the vertices as not visited (For first DFS)
boolean visited[] = new boolean[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Fill vertices in stack according to their finishing
// times
for (int i = 0; i < V; i++)
if (visited[i] == false)
fillOrder(i, visited, stack);
// Create a reversed graph
Graph gr = getTranspose();
// Mark all the vertices as not visited (For second DFS)
for (int i = 0; i < V; i++)
visited[i] = false;
// Now process all vertices in order defined by Stack
while (stack.empty() == false)
{
// Pop a vertex from stack
int v = (int)stack.pop();
// Print Strongly connected component of the popped vertex
if (visited[v] == false)
{
gr.DFSUtil(v, visited);
System.out.println();
}
}
}
// Driver method
public static void main(String args[])
{
// Create a graph given in the above diagram
Graph g = new Graph(5);
g.addEdge(1, 0);
g.addEdge(0, 2);
g.addEdge(2, 1);
g.addEdge(0, 3);
g.addEdge(3, 4);
System.out.println("Following are strongly connected components "+
"in given graph ");
g.printSCCs();
}
}
// This code is contributed by Aakash HasijaFollowing are strongly connected components in given graph
0 1 2
3
4
How to find all bridges in a given graph?
A simple approach is to one by one remove all edges and see if removal of an edge causes disconnected graph. Following are steps of a simple approach for a connected graph.// A C++ program to find bridges in a given undirected graph
#include<iostream>
#include <list>
#define NIL -1
using namespace std;
// A class that represents an undirected graph
class Graph
{
int V; // No. of vertices
list<int> *adj; // A dynamic array of adjacency lists
void bridgeUtil(int v, bool visited[], int disc[], int low[],
int parent[]);
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // to add an edge to graph
void bridge(); // prints all bridges
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v); // Note: the graph is undirected
}
// A recursive function that finds and prints bridges using
// DFS traversal
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
void Graph::bridgeUtil(int u, bool visited[], int disc[],
int low[], int parent[])
{
// A static variable is used for simplicity, we can
// avoid use of static variable by passing a pointer.
static int time = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = *i; // v is current adjacent of u
// If v is not visited yet, then recur for it
if (!visited[v])
{
parent[v] = u;
bridgeUtil(v, visited, disc, low, parent);
// Check if the subtree rooted with v has a
// connection to one of the ancestors of u
low[u] = min(low[u], low[v]);
// If the lowest vertex reachable from subtree
// under v is below u in DFS tree, then u-v
// is a bridge
if (low[v] > disc[u])
cout << u <<" " << v << endl;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = min(low[u], disc[v]);
}
}
// DFS based function to find all bridges. It uses recursive
// function bridgeUtil()
void Graph::bridge()
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
// Initialize parent and visited arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
}
// Call the recursive helper function to find Bridges
// in DFS tree rooted with vertex 'i'
for (int i = 0; i < V; i++)
if (visited[i] == false)
bridgeUtil(i, visited, disc, low, parent);
}
// Driver program to test above function
int main()
{
// Create graphs given in above diagrams
cout << "\nBridges in first graph \n";
Graph g1(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.bridge();
cout << "\nBridges in second graph \n";
Graph g2(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.bridge();
cout << "\nBridges in third graph \n";
Graph g3(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.bridge();
return 0;
}// A Java program to find bridges in a given undirected graph
import java.io.*;
import java.util.*;
import java.util.LinkedList;
// This class represents an undirected graph using
// adjacency list representation
class Graph
{
private int V; // No. of vertices
// Array of lists for Adjacency List Representation
private LinkedList<Integer> adj[];
int time = 0;
static final int NIL = -1;
// Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
// Function to add an edge into the graph
void addEdge(int v, int w)
{
adj[v].add(w); // Add w to v's list.
adj[w].add(v); //Add v to w's list
}
// A recursive function that finds and prints bridges
// using DFS traversal
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
void bridgeUtil(int u, boolean visited[], int disc[],
int low[], int parent[])
{
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
Iterator<Integer> i = adj[u].iterator();
while (i.hasNext())
{
int v = i.next(); // v is current adjacent of u
// If v is not visited yet, then make it a child
// of u in DFS tree and recur for it.
// If v is not visited yet, then recur for it
if (!visited[v])
{
parent[v] = u;
bridgeUtil(v, visited, disc, low, parent);
// Check if the subtree rooted with v has a
// connection to one of the ancestors of u
low[u] = Math.min(low[u], low[v]);
// If the lowest vertex reachable from subtree
// under v is below u in DFS tree, then u-v is
// a bridge
if (low[v] > disc[u])
System.out.println(u+" "+v);
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = Math.min(low[u], disc[v]);
}
}
// DFS based function to find all bridges. It uses recursive
// function bridgeUtil()
void bridge()
{
// Mark all the vertices as not visited
boolean visited[] = new boolean[V];
int disc[] = new int[V];
int low[] = new int[V];
int parent[] = new int[V];
// Initialize parent and visited, and ap(articulation point)
// arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
}
// Call the recursive helper function to find Bridges
// in DFS tree rooted with vertex 'i'
for (int i = 0; i < V; i++)
if (visited[i] == false)
bridgeUtil(i, visited, disc, low, parent);
}
public static void main(String args[])
{
// Create graphs given in above diagrams
System.out.println("Bridges in first graph ");
Graph g1 = new Graph(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.bridge();
System.out.println();
System.out.println("Bridges in Second graph");
Graph g2 = new Graph(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.bridge();
System.out.println();
System.out.println("Bridges in Third graph ");
Graph g3 = new Graph(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.bridge();
}
}Bridges in first graph
3 4
0 3
Bridges in second graph
2 3
1 2
0 1
Bridges in third graph
1 6
Input : mat[][] = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}
Output : 5
Solution - The problem can be easily solved by applying DFS() on each component. In each DFS() call, a component or a sub-graph is visited. We will call DFS on the next un-visited component. The number of calls to DFS() gives the number of connected components. BFS can also be used. {1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}
Pseudo Code
// A utility function to do DFS for a
// 2D boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(int M[][COL], int row, int col,
bool visited[][COL])
{
// These arrays are used to get
// row and column numbers of 8
// neighbours of a given cell
rowNbr[] = { -1, -1, -1, 0, 0, 1, 1, 1 }
colNbr[] = { -1, 0, 1, -1, 1, -1, 0, 1 }
// Mark this cell as visited
visited[row][col] = true
// Recur for all connected neighbours
for (int k = 0; k < 8; ++k)
if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited))
DFS(M, row + rowNbr[k], col + colNbr[k], visited)
}
// The main function that returns
// count of islands in a given boolean
// 2D matrix
int countIslands(int M[][COL])
{
// Make a bool array to mark visited cells.
// Initially all cells are unvisited
bool visited[ROW][COL]
memset(visited, 0, sizeof(visited))
// Initialize count as 0 and
// travese through the all cells of
// given matrix
int count = 0
for (int i = 0; i < ROW; ++i)
for (int j = 0; j < COL; ++j)
// If a cell with value 1 is not
if (M[i][j] && !visited[i][j]) {
// visited yet, then new island found
// Visit all cells in this island.
DFS(M, i, j, visited)
// and increment island count
++count
}
return count
}
// Utility function to check back edge in a directed graph
bool isCyclicUtil(int v, bool visited[], bool *recStack)
{
if(visited[v] == false)
{
// Mark the current node as visited and part of recursion stack
visited[v] = true
recStack[v] = true
// Recur for all the vertices adjacent to this vertex
list < int > :: iterator i
for(i = adj[v].begin(); i != adj[v].end(); ++i)
{
if ( !visited[*i] && isCyclicUtil(*i, visited, recStack) )
return true
else if (recStack[*i])
return true
}
}
recStack[v] = false; // remove the vertex from recursion stack
return false;
}
// Returns true if the graph contains a cycle, else false.
bool isCyclic()
{
// Mark all the vertices as not visited and not part of recursion
// stack
bool *visited = new bool[V]
bool *recStack = new bool[V]
for(int i = 0; i < V; i++)
{
visited[i] = false
recStack[i] = false
}
// Call the recursive helper function to detect cycle in different
// DFS trees
for(int i = 0; i < V; i++)
if (isCyclicUtil(i, visited, recStack))
return true
return false
}
// A recursive function used by topological sort
void topologicalSortUtil(int v, bool visited[],
stack&Stack)
{
// Mark the current node as visited.
visited[v] = true
// Recur for all the vertices adjacent to this vertex
list < int > :: iterator i
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (!visited[*i])
topologicalSortUtil(*i, visited, Stack);
// Push current vertex to stack which stores result
Stack.push(v)
}
// The function to do Topological Sort. It uses recursive
// topologicalSortUtil()
void topologicalSort()
{
stack < int > Stack
// Mark all the vertices as not visited
bool *visited = new bool[V]
for (int i = 0; i < V; i++)
visited[i] = false
// Call the recursive helper function to store Topological
// Sort starting from all vertices one by one
for (int i = 0; i < V; i++)
if (visited[i] == false)
topologicalSortUtil(i, visited, Stack)
// Print contents of stack
// Topological Order
while (Stack.empty() == false)
{
print(Stack.top())
Stack.pop()
}
}
0: Empty cellSo we have to determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index [ i,j ] can rot other fresh orange at indexes [ i-1, j ], [ i+1, j ], [ i, j-1 ], [ i, j+1 ] (up, down, left and right). If it is impossible to rot every orange then simply return -1.
1: Cells have fresh oranges
2: Cells have rotten oranges
Input: arr[][C] = { {2, 1, 0, 2, 1},
{1, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges can become rotten in 2 time frames.
Input: arr[][C] = { {2, 1, 0, 2, 1},
{0, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges cannot be rotten.
1) Create an empty Q.Pseudo Code
2) Find all rotten oranges and enqueue them to Q. Also enqueue a delimiter to indicate the beginning of next time frame.
3) While Q is not empty do following
....3.a) Do following while delimiter in Q is not reached
........ (i) Dequeue an orange from queue, rot all adjacent oranges. While rotting the adjacent, make sure that the time frame is incremented only once. And the time frame is not incremented if there are no adjacent oranges.
....3.b) Dequeue the old delimiter and enqueue a new delimiter. The oranges rotten in the previous time frame lie between the two delimiters.
// This function finds if it is possible to rot all oranges or not.
// If possible, then it returns minimum time required to rot all,
// otherwise returns -1
int rotOranges(int arr[][C])
{
// Create a queue of cells
queue < cell > Q
cell temp
ans = 0
// Store all the cells having rotten orange in first time frame
for (int i=0; i < R; i++)
{
for (int j=0; j < C; j++)
{
if (arr[i][j] == 2)
{
temp.x = i
temp.y = j
Q.push(temp)
}
}
}
// Separate these rotten oranges from the oranges which will rotten
// due the oranges in first time frame using delimiter which is (-1, -1)
temp.x = -1
temp.y = -1
Q.push(temp)
// Process the grid while there are rotten oranges in the Queue
while (!Q.empty())
{
// This flag is used to determine whether even a single fresh
// orange gets rotten due to rotten oranges in current time
// frame so we can increase the count of the required time.
bool flag = false
// Process all the rotten oranges in current time frame.
while (!isdelim(Q.front()))
{
temp = Q.front()
// Check right adjacent cell that if it can be rotten
if (isvalid(temp.x+1, temp.y) && arr[temp.x+1][temp.y] == 1)
{
// if this is the first orange to get rotten, increase
// count and set the flag.
if (!flag) ans++, flag = true
// Make the orange rotten
arr[temp.x+1][temp.y] = 2
// push the adjacent orange to Queue
temp.x++
Q.push(temp)
temp.x-- // Move back to current cell
}
// Check left adjacent cell that if it can be rotten
if (isvalid(temp.x-1, temp.y) && arr[temp.x-1][temp.y] == 1) {
if (!flag) ans++, flag = true
arr[temp.x-1][temp.y] = 2
temp.x--
Q.push(temp) // push this cell to Queue
temp.x++
}
// Check top adjacent cell that if it can be rotten
if (isvalid(temp.x, temp.y+1) && arr[temp.x][temp.y+1] == 1) {
if (!flag) ans++, flag = true
arr[temp.x][temp.y+1] = 2
temp.y++
Q.push(temp) // Push this cell to Queue
temp.y--
}
// Check bottom adjacent cell if it can be rotten
if (isvalid(temp.x, temp.y-1) && arr[temp.x][temp.y-1] == 1) {
if (!flag) ans++, flag = true
arr[temp.x][temp.y-1] = 2
temp.y--
Q.push(temp) // push this cell to Queue
}
Q.pop()
}
// Pop the delimiter
Q.pop()
// If oranges were rotten in current frame than separate the
// rotten oranges using delimiter for the next frame for processing.
if (!Q.empty()) {
temp.x = -1
temp.y = -1
Q.push(temp)
}
// If Queue was empty than no rotten oranges left to process so exit
}
// Return -1 if all arranges could not rot, otherwise -1.
return (checkall(arr))? -1: ans
}
I. 7, 6, 5, 4, 4, 3, 2, 1
II. 6, 6, 6, 6, 3, 3, 2, 2
III. 7, 6, 6, 4, 4, 3, 2, 2
IV. 8, 7, 7, 6, 4, 2, 1, 1
A | I and II |
B | III and IV |
C | IV only |
D | II and IV |